2 câu trả lời
`(x^2-x-2)\sqrt{x+1}=0` (ĐK:`x >= -1`)
`⇔(x^2-2x+x-2)\sqrt{x+1}=0`
`⇔(x-2)(x+1)\sqrt{x+1}=0`
⇔\(\left[ \begin{array}{l}x-2=0\\x+1=0\\\sqrt{x+1}=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2\\x=-1\\x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2(tm)\\x=-1(tm)\end{array} \right.\)
Vậy ptr có 2 nghiệm là: `x=2` hoặc `x=-1`
`(x^2-x-2)\sqrt{x+1}=0`
⇔ $\left[\begin{matrix} x^2-x-2 =0 \\ \sqrt{x+1} =0 \end{matrix}\right.$
⇔ $\left[\begin{matrix} (x^2-2x)+(x-2) =0 \\ x+1 =0 \end{matrix}\right.$
⇔ $\left[\begin{matrix} x(x-2)+(x-2) =0 \\ x=-1 \end{matrix}\right.$
⇔ $\left[\begin{matrix} (x-2)(x+1) =0 \\ x=-1 \end{matrix}\right.$
⇔ $\left[\begin{matrix} x=2 \\ x=-1 \\ x=-1 \end{matrix}\right.$
⇔ $\left[\begin{matrix} x=2 (TM) \\ x=-1 (TM) \end{matrix}\right.$
`S={2;-1}`
vậy phương trình có 2 nghiệm
